Understanding pH and pOH: Calculating the pH of a 2.5x10^-5 M NaOH Solution

In this article, we will explore the basic principles of pH and pOH calculations, particularly focusing on determining the pH of a 2.5x10^-5 M NaOH (sodium hydroxide) solution. We'll also provide a set of useful formulas and explain the concept of autoprotolysis in water.

Key Formulas for pH and pOH Calculations

Here are some essential formulas that will help you solve pH or pOH related questions:

Calculation of the pH of a 2.5x10^-5 M NaOH Solution

Let's start by calculating the pH of a 2.5x10^-5 M NaOH solution:

NaOH → Na OH-

pOH -log[OH-] -log(2.5 x 10-5) 4.6

pH 14 - pOH 14 - 4.6 9.4

Autoprotolysis of Water and pH Calculations

Water undergoes autoprotolysis, producing equal concentrations of hydronium ions (H3O ) and hydroxide ions (OH-):

2H2O ? H3O OH-

The equilibrium constant for this reaction is the ion product of water (Kw):

[K_w [H_3O^ ][OH^-] 10^{-14}]

By taking the negative log of both sides, we can derive the relationship between pH and pOH:

[-log_{10}K_w -log_{10}[H_3O^ ] - log_{10}[OH^-] -log_{10}(10^{-14})]

Since [-log_{10}(10^{-14}) 14], we have:

14 pH pOH

Consistency Check with Basic Solution

For a basic solution, pOH should be greater than 7 (since [OH^-] > [H^ ]). Here, 4.6 is not a valid pOH, indicating a basic solution where:

This result is consistent with a basic solution where pOH is low and pH is high.

Summary

In summary, we have discussed the calculation of pH and pOH, the autoprotolysis of water, and the consistent application of these principles to determine the pH of a basic solution. Understanding these concepts is crucial for accurate analysis in chemical and environmental sciences.